What is the limit as x approaches 0+of x^{-4/(ln(x))} ?

The limit as x approaches 0+of x^{-4/(ln(x))} is 1/(e^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches 0+of x^{-4/(ln(x))}

x \to 0 + lim (x^{- \frac{4}{l n ( x )}})

\frac{1}{e ^{4}}

Decimal Notation

0.01831 \dots

Solution steps

x \to 0 + lim (x^{- \frac{4}{l n ( x )}})

Apply exponent rules

= x \to 0 + lim (e^{- \frac{4}{l n ( x )} l n (x)})

e^{- \frac{4}{l n ( x )} l n (x)} = \frac{1}{e ^{4}}

= x \to 0 + lim (\frac{1}{e ^{4}})

x \to a lim c = c

= \frac{1}{e ^{4}}

x \to 0 - lim (\frac{- 1}{x})

y = lo g_{5} (5 x + 4)

\frac{\partial}{\partial y} (4 - x^{2} - 4 y^{2})

n = 0 \sum \infty 1 0^{n} x^{n}

\int x^{2} e^{10 x} d x

What is the limit as x approaches 0+of x^{-4/(ln(x))} ?
The limit as x approaches 0+of x^{-4/(ln(x))} is 1/(e^4)